Explanation:
Bromine is a chemical element with the symbol Br and atomic number 35. It is the third-lightest halogen, and is a fuming red-brown liquid at room temperature that evaporates readily to form a similarly coloured vapour. Its properties are intermediate between those of chlorine and iodine. Isolated independently by two chemists, Carl Jacob Löwig (in 1825) and Antoine Jérôme Balard (in 1826), its name was derived from the Ancient Greek βρῶμος ("stench"), referring to its sharp and disagreeable smell.
Bromine, 35Br
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958 = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass
= 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water
= 4.7 / 0.88 = 5.34 moles
∴ moles of the solution = total moles in solu - moles of water
= 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
= 53.8 / 0.64 = 84 g/mole
Q: C
moles of urea (NH2)2 CO = mass weight / molar mass
= 4.49 g / 60 g /mol
= 0.07 mol
moles of methanol = mass weight / molar mass
= 39.9 g / 32 g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg
∴Pv(solu) = 84.55 mmHg
Answer:
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